2
$\begingroup$

Let $X\coprod X$ be the disjoint union of $X$ with itself and let there be a commutative square

$$\require{AMScd} \begin{CD} X \coprod X @>{f,g}>> A \\ @V{(i_0, i_1)}VV @VV{p}V \\ X \times I @>>{k}> B \end{CD} $$

where $p$ is a Serre fibration and weak homotopy equivalence. Is it true that this diagram admits a lift $h: X \rightarrow A$ ? (In other words, is $X \times I$ a cylinder object for any $X$?). I know this is true for CW-complexes, but I think it holds for any topological space.

$\endgroup$
  • $\begingroup$ I believe the only fact about CW complexes you would use in proving this is that the inclusion into the cylinder is a cofibration, but this is true for all spaces since it is clear that $X \times \{0\} \cup X \times \{1\}$ is a NDR of $X \times I$. $\endgroup$ – Connor Malin Sep 16 at 18:41
  • $\begingroup$ I wrote something wrong, a need that both the top triangle and bottom triagle commute, in what a read about what you said implies the existence of lift making to top commute, but can we take such homotopy to make the bottom traingle commute? $\endgroup$ – HelloDarkness Sep 16 at 19:04
  • $\begingroup$ Let $A$ be any space and take $p$ to be the unique map to a point. Then $f,g$ can be arbitrary maps and there is no reason why they should be homotopic. Clearly in general there is no reason why a diagonal should exist. $\endgroup$ – Tyrone Sep 16 at 19:06
  • $\begingroup$ In this case $p$ wouldn't be serre fibration. $\endgroup$ – HelloDarkness Sep 16 at 19:07
  • $\begingroup$ Yes, it would be. It would even be a Hurewicz fibration. $\endgroup$ – Tyrone Sep 16 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.